\newproblem{lay:3_suppl_9}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 3.Suppl.9}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Nov. 11th, 2013} \\}{}

  % Problem statement
	Let $T$ be the Vandermonde matrix $T=\begin{pmatrix}1 & a & a^2\\ 1 & b & b^2 \\ 1 & c & c^2\end{pmatrix}$. Use row operations
	to show that $|T|=(b-a)(c-a)(c-b)$
}{
  % Solution
	Let us calculate the determinant of $T$\\
	\begin{tabular}{cl}
	   & $\det\begin{pmatrix}1 & a & a^2\\ 1 & b & b^2 \\ 1 & c & c^2\end{pmatrix}=$ \\
		 \begin{tabular}{l}
		    $\mathbf{r}_2\leftarrow \mathbf{r}_2-\mathbf{r}_1$ \\
				$\mathbf{r}_3\leftarrow \mathbf{r}_3-\mathbf{r}_1$
		 \end{tabular} &
		  $\det\begin{pmatrix}1 & a & a^2\\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2\end{pmatrix}=$ \\
		& $1 \det\begin{pmatrix}b-a & b^2-a^2 \\ c-a & c^2-a^2\end{pmatrix}=$ \\
			 \begin{tabular}{l}
		    $\mathbf{r}_2\leftarrow \frac{1}{b-a}\mathbf{r}_2$ \\
				$\mathbf{r}_3\leftarrow \frac{1}{c-a}\mathbf{r}_3$
		 \end{tabular} 
		& $(b-a)(c-a) \det\begin{pmatrix}1 & \frac{b^2-a^2}{b-a} \\ 1 & \frac{c^2-a^2}{c-a}\end{pmatrix}=$ \\
		& $(b-a)(c-a) \det\begin{pmatrix}1 & b+a \\ 1 & c+a\end{pmatrix}=$ \\
		$\mathbf{r}_2\leftarrow \mathbf{r}_2-\mathbf{r}_1$ & $(b-a)(c-a) \det\begin{pmatrix}1 & b+a \\ 0 & c-b\end{pmatrix}=$ \\
		 & $(b-a)(c-a) 1\det\begin{pmatrix}c-b\end{pmatrix}=$ \\
		 & $(b-a)(c-a)(c-b)$ \\
	\end{tabular}
}
\useproblem{lay:3_suppl_9}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
